Rewrite the function by completing the square. $g(x)= 4 x^{2} -16 x +7$ $g(x)=$
Solution: $\begin{aligned} g(x)&=4 x^2 -16 x +7 \\\\ &=4 \left(x^2 -4 x\right) +7 \end{aligned}$ Now we want to complete $x^2 -4 x$ into a perfect square. To do that, we should add $\left(\dfrac{{-4}}{2}\right)^2={4}$ to it: $x^2{-4}x+{4}=\left(x -2\right)^2$ We add ${4}$ inside the parentheses, and subtract ${4}\cdot{4}$ outside them, to keep the expression equivalent. $\begin{aligned} &\phantom{=}{4} \left(x^2 -4 x\right) +7 \\\\ &={4}\left(x^2 -4 x+{4}\right) +7 -{4}\cdot{4} \\\\ &=4 \left(x -2\right)^2 +7 -16 \\\\ &=4 \left(x -2\right)^2 -9 \end{aligned}$ In conclusion, the function after completing the square is written as: $g(x)=4 \left(x -2\right)^2 -9$